From: CCIE 19999 (ccie@iprimus.com.au)
Date: Wed Aug 15 2007 - 23:48:29 ART
The IPV6 addresses are written in HEX.
First step is to write each segment into full HEX digits, so you get
2001:141:1:12::/64 - 2001:0141:0001:0012:: and
2001:141:1:25::/64 - 2001:0141:0001:0025::
The first 3 segments are equal or the 48 bits(16x3), so will concentrate on
4th segment, which are
0012 and
0025
The first 2 hex digits are the same, that is another 8 bits and we are left
with
12 and
25
Which are in binary as:
0001 0010 and
0010 0101
And 2 bits are same, so that's another 2 bits
If add all the common bits now 48 + 8 + 2 = 58
HTH
Shine
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
ccie1101
Sent: Wednesday, 15 August 2007 3:47 PM
To: Cisco certification
Subject: Re : IPv6 Summarization ...
Hi Experts,
How do you summarize between 2001:141:1:12::/64 and 2001:141:1:25::/64
? I don't seem to be able to get the answer 2001:141:1::/58.
Is the summarization the same as in ipv4 but of course each octet has a
value of 16 ? Coz i am doing it as in ipv4 but it does not come
to 2001:141:1::/58 ....
Pls advice on the way to do this ...
Thank you,
Cheers,
ccie1101.
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